I'm pretty sure the method is to figure out some always positive \(f(n)\) such that \(f(n) - \frac{1}{(2n+1)^2} > f(n+1)\), so we can induct it on \(\sum_{i=1}^n \frac{1}{(2i+1)^2} < \frac{1}{4} - f(n)\).

Note that the second condition is superfluous, because we can choose a different starting point. For example, if we take \( b = 2 \), then with \( g(n) = \frac{n+2}{(2n+1)^2} \), we get that \[ g(n) - g(n+1) > \frac{ 1}{ (2n+1) } ^2 \text{ for } n \geq 1 .\]